 to the lecture number 27 of the course, Quantum Mechanics and Molecular Spectroscopy. In the last class, we were looking at the rotational Hamiltonian. The rotational Hamiltonian rotation is given by L square by 2 mu R e square. I have used several, you know R e is identically equal to R naught is odd equal to R. So, that is nothing but the length of the A B bond equilibrium length. So, that is R naught R R e and this is rigid that means it is not changing. So, the bond of A B is fixed length that is also. So, this gives you what is known as rigid rotor approximation. Now, if that is the case this can slightly be written as R square by sorry L square by 2 I e where I e is equal to mu R e square and this is called moment of inertia. And the operator L square is the total angular momentum operator. Now, this value will be equal to L square operator actually L square should be an operator. This will be equal to minus h bar square into 1 over sin theta d by d theta sin theta d by d theta plus 1 over sin square theta d square by d phi square where theta and phi are angular coordinates as part of spherical polar coordinates. Now, there is also something called L z operator that will be equal to z component of the angular momentum. Now, it turns out that the solutions angular momentum problem. By the way, one can imagine this as particle on a surface of a sphere or it can also be is called as particle central force problem. Then what we have is the L square operator will have wave function y j m of theta and phi equals to h bar square j into j plus 1 y j m theta and phi and L z operator acting on y j m theta and phi will give you m h bar y j m theta and phi. Now, using this one can write h rotation is equal to h rotation y j m theta and phi will be equal to. Now, h rotation had 2 1 over 2 i as a additional this is. So, if you go back h rotation was equal to L square by 2 i e. So, we already know the wave functions of this L square operator. So, only you multiply by this 2 i 1 over 2 i. So, that will be equal to h bar square by 2 i e j into j plus 1 y j m theta and phi. So, now you can look at your. So, E rotation will now we depend on 2 quantum numbers j and m that is equal to h bar square by 2 i e j into j plus 1. Even though the wave function depends on m and j the energy depends only on j. So, that is your E rotation. This E rotation is all sometimes also written as rotation j m is equal to h bar square by 2 i e j into j plus 1 or equal to h times b e j into j plus 1 where b e is given by h by 8 pi square i e and this is called rotational constant. So, now we know the wave functions and we know the energies. Now, let us go back to the transition moment integral. The transition moment integral is given by f mu epsilon i. Now, but your mu itself is written as mu is written as mu naught plus T mu by T r, r is equal to 0 into r plus T square mu by T r square evaluated r naught into r square plus etcetera, etcetera whole thing epsilon to i. Now, if you take rigid rotor T mi, if we have rigid rotor then of course, the r naught r naught is a fixed value. Therefore, the derivatives with respect to r will not exist. Therefore, for rigid rotor T mi for rigid rotor will be equal to f because this will go to 0 and this will also go to 0 all the other terms will go to 0 what you have is mu naught where mu naught is called permanent dipole moment. Now, your T mi is given by this. So, you have to have mu naught. So, under rigid rotor approximation for rigid rotor if T mi has to be non-zero T mi that is equal to f mu naught epsilon i this has to be non-zero then mu naught has to be non-zero that is a initial if that is a minimum condition. If mu naught becomes 0 of course, this entire integral becomes 0. So, the mu naught must be not equal to 0. That means, you can only have rotational transitions mu naught is not equal to 0. That means, the permanent dipole moment of molecules must be non-zero to record rotational spectrum ok. So, T mi equals to integral f mu naught dot epsilon. So, for recording rotational spectrum transformation is the permanent dipole moment must be must not be 0. So, only mu naught is not equal to 0 transitions will be allowed ok. So, for example, if you have a ab diatomic molecule where a is not equal to b then this molecule will have permanent dipole moment mu naught is not equal to 0. But if you have a homo diatomics like a, a ok then of course, mu is equal to 0 mu naught is equal to 0. So, which means, you cannot record rotational spectrum is not possible ok. So, one of the selection rules for the recording rotational spectrum is that the transient moment integral will go to 0 when mu naught becomes 0. So, we can only record rotational spectra of molecules that are that have permanent dipole moment ok. Now, let us look at the T mi little more carefully T mi equals to f mu naught and one thing that you must remember that when I have this that means, if you take a molecule ab and your dipole moment is in this direction let us suppose ok. So, this is delta plus and delta minus then the dipole moment will be in that direction. So, the propagation should be in this direction and the electric field should be. So, your e ok should be along the dipole moment ok that is a selection rule. So, this says that the electric field should oscillate along the because you are taking a dot product ok. Of course, you can come at an angle because you can always have a you can always have the projection along that axis. However, if you have a and b and if a direction of propagation is like this and your electric field is like this and your dipole moment is like this ok. So, dipole moment is along let us say x axis ok and the propagation of light is along x axis then your electric field will be along z axis let us assume. So, this is my z axis then of course, x dot z will be equal to 0. Therefore, the transients will not take place ok. So, it is not only that you know the mu naught must be 0, but the electric field must be aligned along the permanent dipole moment to have the transients that is the this TMI ok. So, now once ok I assume that that becomes non-zero ok that means, the molecule has dipole moment. So, two things one is molecule has permanent dipole moment, electric field vector of the light has non-zero projection along the dipole moment. Only then the t a transient moment integral can become non-zero. So, TMI will be equal to f mu naught dot epsilon dot. So, I am only looking at this part, this part needs two conditions one is the dipole moment must be non-zero or the permanent dipole moment must be non-zero and the electric field vector must be aligned or projected have a projection along the dipole moment ok. If these two conditions are satisfied then we can look at the following the f and i ok. So, these are what are known as necessary conditions. So, these two recording necessary conditions without these conditions of course, one cannot record the rotational spectrum ok. Now, if you are know what we have now we have TMI is equal to f mu naught dot epsilon. So, we already taken care we have already taken care of this. Now, let us look at the wave functions. Now, your wave function i will be equal to now this will be the total wave function i. So, what will the total wave function i will consist of that will consist of electronic rot psi E L and the nuclear part while I will call it as a chi nuclear ok. And similarly there is a notation always the ground state is given by double dash and the excited state is given by single dash. So, f is equal to psi electronic ok. Now, the electronic is the this comes from the electronic. So, what is electronic H electronic psi electronic is E electronic psi electronic. So, this is the Newtonian that one needs to solve and this H electronic wave function will H electronic Hamiltonian will consist of Ke of electrons electrons and nuclear and Pe of electrons repulsion either this is attraction Pe of nuclear this is repulsion ok. Now, so this is the H electronic that you have to use different techniques you have to do a quantum chemistry to solve that, but in this course we will assume that we know that ok. Now, the chi nuclear will have. So, the chi nuclear will have two parts one is the what is what is known as phi nuclear or phi internal into phi internal into the other thing is the translation ok. So, your i will now become psi electronic into phi internal into phi translation and your f will be equal to psi electronic phi internal translation ok. So, now let us write the transition moment integral. So, your TMI will be equal to psi electronic phi translation phi internal dash mu naught psi electronic trans phi internal. Now, if you are looking at rigid rotor approximation let us think of a molecule AB separated by distance r naught or re whatever sum and then it is rotating ok. So, what will happen to electrons nothing will happen to the electronic part of the wave function remains unchanged. So, psi EL remains unchanged. So, which means psi EL double prime should be equal to psi EL single prime this is equal to psi EL. Now, if it is rotating and it is translating. So, the rotation will not will assume that rotation is not going to affect the translation rotational degrees of freedom and transnational degrees of freedom can be separated out. So, then what happens is that phi trans prime will be equal to phi trans when you have the electronic degrees of freedom frozen your dipole moment is not going to change ok. Similarly, when you have translation if the AB molecule is translating its dipole moment is not going to change. So, I can write read my TMI as psi electronic ok. So, this is now 3 integrals is not it psi electronic phi trans phi internal mu naught epsilon psi electronic phi trans phi internal ok this is single prime and this is double prime. So, that is my. So, since the psi electronic and psi trans these two quantities are not going to depend on the this then I can bring them out that means I will have psi electronic phi trans and psi electronic phi trans and what I will have is the phi ok. Now, if you are electronic wave function and translation wave function are normalized then this will go to 0 sorry this will go to 1 ok wave functions are therefore your TMI will become phi internal mu naught epsilon phi ok. So, for rotation the TMI will now become your phi internal is nothing but your rotation wave function. So, I will call it as a phi rotation in the excited state ok mu naught epsilon phi rotation in the ground state wave functions. Now, we know this will be nothing but if you take rigid rotor approximation the phi rotation will nothing but y j m j prime m prime theta and phi mu naught epsilon y j double prime m double prime theta and phi. So, that is going to be your TMI ok. Your transition moment integral can now be written in terms of spherical, spherical harmonics. So, transition moment integral can be written y j m theta and phi which is nothing but your spherical ok. I am going to stop here and we will continue in the next lecture. Thank you. .